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如何获取HTTP post请求以获取响应iPhone应用程序的JSON对象?

如何解决《如何获取HTTPpost请求以获取响应iPhone应用程序的JSON对象?》经验,为你挑选了1个好方法。

我在服务器上运行的Web服务以XML格式或JSON格式返回数据.我想要请求JSON格式但使用HTTP Post方法.

任何帮助非常感谢.

提前致谢.



1> Amit Vaghela..:

这是用于JSON post请求的代码,TouchJSON Framework用于解析JSON,感谢'schwa'.

NSArray *keys = [NSArray arrayWithObjects:@"username", @"password", @"preference", @"uid", nil];
NSArray *objects = [NSArray arrayWithObjects:@"accuser", @"accpass", @"abc_region", @"100", nil];
NSDictionary *theRequestDictionary = [NSDictionary dictionaryWithObjects:objects forKeys:keys];

NSURL *theURL = [NSURL URLWithString:@"http://url.com/request.php"];
NSMutableURLRequest *theRequest = [NSMutableURLRequest requestWithURL:theURL cachePolicy:NSURLRequestReloadIgnoringCacheData timeoutInterval:10.0f];
[theRequest setHTTPMethod:@"POST"];

[theRequest setValue:@"application/json-rpc" forHTTPHeaderField:@"Content-Type"];
NSString *theBodyString = [[CJSONSerializer serializer] serializeDictionary:theRequestDictionary];
NSLog(@"%@", theBodyString);
NSData *theBodyData = [theBodyString dataUsingEncoding:NSUTF8StringEncoding];
// NSLog(@"%@", theBodyData);
[theRequest setHTTPBody:theBodyData];

NSURLResponse *theResponse = NULL;
NSError *theError = NULL;
NSData *theResponseData = [NSURLConnection sendSynchronousRequest:theRequest returningResponse:&theResponse error:&theError];
NSString *theResponseString = [[[NSString alloc] initWithData:theResponseData encoding:NSUTF8StringEncoding] autorelease];
NSLog(theResponseString);
NSDictionary *theResponseDictionary = [[CJSONDeserializer deserializer] deserialize:theResponseString];
NSLog(@"%@", theResponseDictionary);
NSString *theGreeting = [theResponseDictionary objectForKey:@"greeting"];
[self setValue:theGreeting forKey:@"greeting"];

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